Question

cosX-4sinx=1 then sinx+4cosx
$\cos(x) - 4 \sin(x) = 1 \: \\ then \: \sin(x) + 4cos(x)$
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Answer 1

Answer:

sinx + 4 cosx = ± 4

Step-by-step explanation:

Formula used:

$(a+b)^2=a^2+b^2+2ab$

$(a-b)^2=a^2+b^2-2ab$

$sin^2\theta=1-cos^2\theta$

$cos^2\theta=1-sin^2\theta$

$cosx-4sinx=1$

squaring on both sides

$(cosx-4sinx)^2=1^2$

$cos^2x+16sin^2x-8\:cosx\:sinx=1$

$(1-sin^2x)+16(1-cos^2x)-8\:cosx\:sinx=1$

$1-sin^2x+16-16cos^2x-8\:cosx\:sinx=1$

$-sin^2x-16cos^2x-8\:cosx\:sinx=-16$

$sin^2x+16cos^2x+8\:cosx\:sinx=16$

$sin^2x+(4cosx)^2+8\:sinx\:cosx=16$

$(sinx+4cosx)^2=4^2$

Taking square root on both sides

sinx + 4 cosx = ±4

Answer 2

Answer:

cosX-4sinx=1 then

Sinx + 4 Cosx = ±4

Step-by-step explanation:

Cox - 4 Sinx = 1

=> Cosx = 1 + 4Sinx

Squaring both sides

=> Cos²x = 1 + 16Sin²x + 8Sinx

=> 0 = 1 - Cos²x + 16Sin²x + 8Sinx

=> 0 = Sin²x + 16Sin²x + 8Sinx

=>  0 = 17Sin²x + 8Sinx

=> Sinx( 17Sinx + 8) = 0

Sinx = 0   or Sinx = -8/17

Case 1

Sinx = 0

Cosx =  1 + 4Sinx  = 1

Sinx + 4cosx = 0 + 4(±1) = ±4

Case 2

Sinx = -8/17

Cosx = 1 + 4 Sinx = 1 + 4(-8/17) = -15/17

Sinx + 4 Cosx =  -8/17 + 4(-15/17)

= -8/17 - 60/17

= -68/17

= -4

Combining case 1 & Case 2

Sinx + 4 Cosx = ±4